On certain inequalities for the prime counting function – Part III

: As a continuation of [10] and [11], we offer some new inequalities for the prime counting function π ( x ) . Particularly, a multiplicative analogue of the Hardy–Littlewood conjecture is provided. Improvements of the converse of Landau’s inequality are given. Some results on π ( p 2 n ) are offered, p n denoting the n -th prime number. Results on π ( π ( x )) are also considered


Introduction
Let π(x) denote the number of primes ≤ x, where x ≥ 1 is a positive integer. In Parts I and II [10,11] we have proved some inequalities of a new type for π(x).

Main results
The following auxiliary results will be used: For references to this results, see [10,11].
A simple computation (which we omit here) gives the second derivative of this function: Thus the function f is concave, which gives for any x, y ≥ 23.
Now, using the right side of (9), and by (13) we get On the other hand, by the left side of (9) we get

Now we have to considered the inequality
which can be written , after elementary computations as log(x + y) > 5.75 . . . ; i.e., x + y > e 5.75... ≈ 317.34 . . . (11) is true for x, y ≥ 67 and x + y ≥ 318. A computer verification shows that (11) is true also (with strict inequality) for 67 ≤ y ≤ x and x + y ≤ 317. Therefore (11) is valid with strict inequality for any x, y ≥ 67.
Since π(x) ≤ π(x + y), it will be sufficient to consider the inequality a computer verification shows the exceptions listed in Theorem 1, as well the equality cases.

Remark 1.
By letting x = p n , the n-th prime number, we get that for n ̸ = 3 one has Particularly, as 3 2 > √ 2, we get that for n ̸ = 3 which was an open problem stated in [4].
The following result gives multiplicative analogues of the Hardy-Littlewood conjecture.
This is valid, if π(x) ≥ 3 and π(y) ≥ 3; i.e., when x, y ≥ 7. As we supposed x, y ≥ 3; the cases of exceptions can be verified, and also the cases of equality can be verified, and also the cases of equality can be easily shown.
The second one is the consequence of 2 √ ab ≤ a + b for a = π(y), b = π(x) x/y .
In our opinion, even with 1 in place of (4) we have a difficult open problem. We have the following res Proof. First we prove the following auxiliary result.
Indeed, R. E. Dressler et al. (see [12]) proved that p 2 n+1 ≤ 2p 2 n for n > 4. A similar argument can be applied for the proof of (40). This is based on the Rosser-Schoenfeld inequalities p n < n(log n + log log n − 1 2 ) for n ≥ 20 and p n > n(log n + log log n − 3 2 ) for n ≥ 2.